Question

Let a₁, a₂, a₃,…. be an A.P. If
\(\begin{array}{l} \displaystyle\sum\limits_{r=1}^\infty\frac{a_r}{2^r}=4,\end{array}\)
then 4a₂ is equal to ________.

Answer 1

Correct Answer: 16

Given an arithmetic progression (A.P.) \(a_1,a_2,a_3,\ldots\), where the sum of terms in the form \(\displaystyle\sum\limits_{r=1}^\infty\frac{a_r}{2^r}=4\), we need to find \(4a_2\).

In an A.P., each term \(a_r\) can be expressed as \(a_r=a_1+(r-1)d\), where \(a_1\) is the first term and \(d\) is the common difference. Therefore:

\(\displaystyle\sum_{r=1}^\infty \frac{a_1+(r-1)d}{2^r}=4\)

We can split the sum into two parts:

\(\displaystyle\sum_{r=1}^\infty \frac{a_1}{2^r} + \sum_{r=1}^\infty \frac{(r-1)d}{2^r} = 4\)

1. The first sum is a geometric series:

\(\displaystyle\sum_{r=1}^\infty \frac{a_1}{2^r} = a_1 \sum_{r=1}^\infty \frac{1}{2^r}\)

The sum of an infinite geometric series with first term \(\frac{1}{2}\) and common ratio \(\frac{1}{2}\) is:

\(\sum_{r=1}^\infty \frac{1}{2^r} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1\)

Thus, \(\sum_{r=1}^\infty \frac{a_1}{2^r} = a_1\).

2. For the second sum, consider:

\(\sum_{r=1}^\infty \frac{(r-1)d}{2^r} = d \sum_{r=1}^\infty \frac{r-1}{2^r}\)

Decompose using the known series formula for \(\sum_{r=1}^\infty \frac{r}{2^r}\) which is \(\frac{1}{(1-\frac{1}{2})^2}=4\):

\(\sum_{r=1}^\infty \frac{r}{2^r} = 2\), so \(\sum_{r=1}^\infty \frac{1}{2^r}=1\)

\(\Rightarrow \sum_{r=1}^\infty \frac{r-1}{2^r} = \sum_{r=1}^\infty \frac{r}{2^r} - \sum_{r=1}^\infty \frac{1}{2^r} = 2-1 = 1\), thus \(d \cdot 1 = d\).

Substituting back, we have:

\(a_1 + d = 4\)

The task is to find \(4a_2 = 4(a_1 + d)\). From \(a_1 + d = 4\), it follows:

\(4(a_1 + d) = 4 \times 4 = 16\)

Therefore, \(4a_2=16\), which fits within the provided range \([16,16]\).

Answer 2

Given, 
\(\begin{array}{l} S=\frac{a_1}{2}+\frac{a_2}{2^2}+\frac{a_3}{2^3}+\frac{a_4}{2^4}+\cdots\infty\end{array}\)
\(\begin{array}{l} \frac{\frac{1}{2}S=~~\frac{a_1}{2^2}+\frac{a_2}{2^3}+\cdots\infty}{\frac{S}{2}~~=\frac{a_1}{2}+\frac{\left(a_2+a_1\right)}{2^2}+\frac{\left(a_3+a_2\right)}{2^3}}+\cdots\infty\end{array}\)
\(\begin{array}{l} \Rightarrow\ \frac{S}{2}=\frac{a_1}{2}+\frac{d}{2}\end{array}\)
⇒ a₁ + d = a₂ = 4 ⇒ 4a₂ = 16